Find the Maclaurin series for f (x) using the definition of the Maclaurin series. $f ( x ) = x \cos ( 4 x )$
A) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } 5 ^ { 2 n } x ^ { 2 n + 1 } } { ( 2 n ) ! }$
B) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } 5 ^ { 2 n } x ^ { 2 n + 1 } } { n ! }$
C) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n + 1 } 5 ^ { 2 n } x ^ { 2 n + 1 } } { ( 2 n ) ! }$
D) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } 5 ^ { n } x ^ { 2 n + 1 } } { ( 2 n ) ! }$
E) $\sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } 5 ^ { 2 n } x ^ { 2 n } } { ( 2 n ) ! }$

Question

Quizplus · Accepted Answer

The Maclaurin series for f(x) is given by:$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n$$where $f^{(n)}(0)$ is the nth derivative of f(x) evaluated at x = 0.For $f(x) = x \cos(4x)$, we have:$$f'(x) = \cos(4x) - 4x \sin(4x)$$$$f''(x) = -4\sin(4x) - 16x \cos(4x)$$$$f'''(x) = -16\cos(4x) + 64x \sin(4x)$$$$f^{(4)}(x) = 64\sin(4x) + 256x \cos(4x)$$and so on.Evaluating these derivatives at x = 0, we get:$$f(0) = 0$$$$f'(0) = 1$$$$f''(0) = -4$$$$f'''(0) = 0$$$$f^{(4)}(0) = 64$$and so on.Therefore, the Maclaurin series for $f(x) = x \cos(4x)$ is:$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n = x - \frac{4}{2!} x^2 + \frac{64}{4!} x^4 - \cdots = \sum_{n=0}^\infty \frac{(-1)^n 5^{2n} x^{2n+1}}{(2n)!}$$

Find the Maclaurin Series for F (X) Using the Definition f(x)=xcos⁡(4x)f ( x ) = x \cos ( 4 x )f(x)=xcos(4x)

Find the Maclaurin Series for F (X) Using the Definition $f ( x ) = x \cos ( 4 x )$